Draw out the conclusion that the product of two numbers is a bit less than the square of the average ...

and the error is the square of the distance to the average.

60*66 is roughly 63*63, and the error is $(63-60)^2$

So $60*66=63^2-3^2=3960$

So a product, any product, every product, is a square minus an error.

Now let's look at $x^2+12x+27.$ This should be a square minus the error. Let's suppose it's $(x+a)^2.$ That's $x^2+2ax+a^2.$ We use whatever value of a make the middle term work, so we want $2ax=12x.$ So let's set a=6.

That gives us $(x+6)^2=x^2+12x+36,$ which is too big by 9. That's the square of the amount that a is the wrong thing to use. The square root of 9 is 3. That means we should use a+3 and a-3, which is 6+3 and 6-3, 9 and 3.

This seems a long drag, but it starts with some really useful
number familiarisation. By playing with squares and seeing how
products are more-or-less close to the square of the average,
students can get a feel for numbers. We can even demonstrate
by drawing a square, and seeing how the area varies if we
increase one side and decrease the other. The area is nearly
constant, and the error is itself a square.

Diagrams to follow if someone asks.

This method can even lead to the standard formula. Writing
that slightly differently than usual, the solutions are -a+e
and -a-e, where a is half the coefficient of x and e
is the square root of the error. If the error is negative you
have no real solutions.